If $f(x)=mlogx+nx^2 + x $ has extreme values at x= 1 and x = -2, then the values of m and n respectively are :

$-2, \frac{1}{2}$

The correct answer is Option (4) → $-2, \frac{1}{2}$

$\frac{d(f(x))}{dy}=\frac{d(m\log x+nx^2+x)}{dy}$

$=\frac{m}{x}+2nx+1$

$f'(1)=m+2n+1=0$

$⇒m+2n=-1$   ....(1)

$f'(-2)=\frac{-m}{2}-4n+1=0$

$-m-8n=-2$

$m+8n=2$   .....(2)

Eq. (2) - 4 × Eq. (1)

$-3m=6$

$m=-\frac{1}{2},n=-2$